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Re: A solid metal cylinder was dropped into a cylindrical glass [#permalink]
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Fixed . it was \(\pi\)

many thanks.

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Re: A solid metal cylinder was dropped into a cylindrical glass [#permalink]
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volume of water in the glass cylinder = πr^2h1 = 200π; r^2 = 200/h1

volume of solid metal cylinder = π*3*3*7 = 63π
resulting water and metal volume in glass cylinder = 200π + 63π = 263π
if the water level reached up by h2 height, the filled volume in the glass cylinder = πr^2h2 = 263π;
h2 = 263/r^2 = (263/200)h1
percentage of water level rises in glass cylinder = (h2-h1)/h1 * 100

h2-h1)/h1 * 100
[(263/200)h1 - h1]/h1 * 100
63/200 * 100 = 31.5
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Re: A solid metal cylinder was dropped into a cylindrical glass [#permalink]
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Volume of cylinder = π*r^2*h
Thus, 7*9*π = 63π
The water level in the glass increases from 200π by 200π+63π = 263π.
Percent increases = 263π-200π/200π * 100 = 63/200*100 = 31.5.

Originally posted by sukrut96 on 17 May 2020, 21:11.
Last edited by sukrut96 on 08 Jul 2020, 00:59, edited 1 time in total.
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Re: A solid metal cylinder was dropped into a cylindrical glass [#permalink]
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Awesomeee


vndnjn wrote:
volume of water in the glass cylinder = πr^2h1 = 200π; r^2 = 200/h1

volume of solid metal cylinder = π*3*3*7 = 63π
resulting water and metal volume in glass cylinder = 200π + 63π = 263π
if the water level reached up by h2 height, the filled volume in the glass cylinder = πr^2h2 = 263π;
h2 = 263/r^2 = (263/200)h1
percentage of water level rises in glass cylinder = (h2-h1)/h1 * 100

h2-h1)/h1 * 100
[(263/200)h1 - h1]/h1 * 100
63/200 * 100 = 31.5
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Re: A solid metal cylinder was dropped into a cylindrical glass [#permalink]
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[quote="Carcass"]Awesomeee

Thank You Carcass :-D
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Re: A solid metal cylinder was dropped into a cylindrical glass [#permalink]
i have a doubt
is it possible any of these questions from the entire challenge are from the two free powerprep tests?
because i have'nt given them yet and would want to steer clear of them
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Re: A solid metal cylinder was dropped into a cylindrical glass [#permalink]
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sukrut96 wrote:
Area of cylinder = π*r^2*h
Thus, 7*9*π = 63π
The water level in the glass increases from 200π by 200π+63π = 263π.
Percent increases = 263π-200π/200π * 100 = 63/200*100 = 31.5.


Nice explanation, the only error is π*r^2*h is the formula of Volume of cylinder and not the Area

Originally posted by Farina on 08 Jul 2020, 00:05.
Last edited by Farina on 08 Jul 2020, 07:00, edited 2 times in total.
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Re: A solid metal cylinder was dropped into a cylindrical glass [#permalink]
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tanvi1998 wrote:
i have a doubt
is it possible any of these questions from the entire challenge are from the two free powerprep tests?
because i have'nt given them yet and would want to steer clear of them


No,

the source is mGRE 100%

All the questions from powerprep are here https://gre.myprepclub.com/forum/gre-power ... html#p6284

See also Free GRE Materials - Where to get it!

https://gre.myprepclub.com/forum/free-gmat ... tml#p53911

Regards
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Re: A solid metal cylinder was dropped into a cylindrical glass [#permalink]
Yes the answer is 63/200 as mentioned by others.

Adding to that: the answer can be obtained by simply doing percentage change in volume

(263pi - 200pi)/200pi = 63/200

That is, the percentage change in volume is the percentage change in height.

Why is this the case?

B/c percentage change in volume is happening solely due to percentage change in the height, since the height of the water level is the only variable changing when the metal cylinder is added to the water.
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Re: A solid metal cylinder was dropped into a cylindrical glass [#permalink]
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Re: A solid metal cylinder was dropped into a cylindrical glass [#permalink]
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